| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring direct application of standard formulas: (a) area = ½ab sin C with all values given, and (b) cosine rule with known sides and included angle. Both are routine calculations with no problem-solving or conceptual challenge beyond formula recall. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| 1(a) | Area = \(\frac{1}{2} \times 5 \times 4.8 \times \sin 30°\) = 6 cm² | M1, A1 |
| 1(b) | \(AB^2 = 5^2 + 4.8^2 - 2 \times 5 \times 4.8 \cos 30°\) = 25 + 23.04 − 41.569 = 6.4707 ⟹ \(AB = \sqrt{6.47...} = 2.5437\) = 2.54 cm to 3 sf | M1, m1, A1 |
1(a) | Area = $\frac{1}{2} \times 5 \times 4.8 \times \sin 30°$ = 6 cm² | M1, A1 | Use of $\frac{1}{2}ab\sin C$ OE; Condone absent cm²; [Note: Calculator set in wrong mode, penalise only once on the paper.] | Accept 'better' than 2.54; Condone absent cm |
1(b) | $AB^2 = 5^2 + 4.8^2 - 2 \times 5 \times 4.8 \cos 30°$ = 25 + 23.04 − 41.569 = 6.4707 ⟹ $AB = \sqrt{6.47...} = 2.5437$ = 2.54 cm to 3 sf | M1, m1, A1 | RHS of cosine rule used; Correct order of evaluation; Accept 'better' than 2.54; Condone absent cm |1 The diagram shows a triangle $A B C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{37627fc4-a90b-4f3b-9b10-0a9e200f8485-2_423_707_612_657}
The lengths of $A C$ and $B C$ are 5 cm and 4.8 cm respectively.\\
The size of the angle $B C A$ is $30 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the area of the triangle $A B C$.
\item Calculate the length of $A B$, giving your answer to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q1 [5]}}